Solutions to oddnumbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 3


 Thomas Kelly
 1 years ago
 Views:
Transcription
1 Solutions to oddnumbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 3 3. (a) Yes; (b) No; (c) No; (d) No; (e) Yes; (f) Yes; (g) Yes; (h) No; (i) Yes. Comments: (a) is the additive group of the Boolean ring; (e) is a subgroup of the multiplicative group of R; and (f) is a subgroup of the multiplicative group of C (apply the subgroup test). (b) This example satisfies the closure and associative laws. Is there a set E such that A E = A for every subset A of X? Yes; in fact the only such subset is the empty set (for /0 E = /0 implies that E = /0). But now, as long as X is not itself empty, we see that the inverse law holds: there is no set A such that X A = /0, since X A is at least as big as X. So this example is not a group as long as X /0. [If it happens that X = /0, then P(X) has just one element, namely /0, and we have the trivial group with one element.] (c) The associative law fails. For example, if A = B = C = {}, then A \ (B \C) = {} \ /0 = {}, (A \ B) \C = /0 \ {} = /0. (d) The inverse law fails: the identity element is, and 0 has no inverse. The proof of (g) by direct calculation is quite difficult. A trick makes it easier. Use the hyperbolic tangent function tanh(x) = (e x e x )/(e x + e x ). This function is strictly increasing and maps R onto the interval (,); and it satisfies the equation tanh(x + y) = tanhx + tanhy + tanhxtanhy. So it is an isomorphism from the additive group (R,+) to (G, ) (in the case c = ); this structure, being isomorphic to a group, must itself be a group. For an arbitrary value of c, simply rescale (use the function ctanhx) Call the matrices I,A,B,C,D,E. Construct a Cayley table. (This involves a fair amount of work.) From the Cayley table we read off the closure law, the identity law (I is the identity), and the inverse law. The associative law holds because matrix multiplication is associative. So the matrices do form a group. It is not abelian: again, two noncommuting matrices can be found from the Cayley table. (For example, AC = D but CA = E.) 3.5. U(R) is infinite. For ( + 2)( + 2) =, so + 2 is a unit. Then all its powers are units, and clearly they are all distinct (a) If gh = hg then ghgh = gghh, and conversely (cancelling g from the left and h from the right). (b) Since g h = (hg), the result is clear. (c) Suppose that (gh) n = g n h n holds for n = m,m +,m + 2. The equatins for n = m,m + give g n+ h n+ = (gh) n gh = g n h n gh.
2 Cancelling g n from the left and h from the right, we see that gh n = h n g, that is, g commutes with h n. Simimlarly, the equations for m = n+,n+2 show that g commutes witth h n+. So g commutes with h n+ h n = h, as required. (The last step can be done by direct calculation, or by showing that the set of elements which commute with g (the socalled centraliser of g) is a subgroup.) 3.9. We are given (G0) and (G) and half of each of the conditions (G2) and (G3), and have to prove the other half. That is, we must show that g e = g (in (b)) and g h = e (in (c)). We prove the second of these things first. Given g G, let h G be as in (c). Also by (c), there exists k G with k h = e. Now we have (k h) (g h) = e (g h) = g h, k ((h g) h) = k (e h) = k h = e, and these two expressions are equal by the Associative Law. Now, if h is as in (c), we have g e = g (h g) = (g h) g = e g = g. 3.. Recall that, if n > 0, then g n is defined by induction: g = g and g n+ + g n g. Also, g 0 = and g m = (g m ) for m > 0. Alternatively, if n > 0, then g n is the product of n factors equal to g, and if n < 0, it is the product of n factors equal to g. The last form is the most convenient. (Here we implicitly used that (g n ) = (g ) n. This holds because g n (g ) n is the product of n factors g followed by n factors g ; everything cancels, leaving the identity.) To prove that g m+n = g m g n, there are nine different cases to consider, according to whether m and n are positive, zero or negative. If one or other of them is zero, the result is easy: for example, g m+0 = g m = g m = g m g 0. This leaves four cases. If m,n > 0, then g m g n is the product of m factors g followed by the product of n factors g, which is the product of m + n factors g, that is, g m+n. Suppose that m is positive and n negative, say m = r. Then g m g n is the product of m factors g followed by r factors g. If m r, then r of the gs cancel all the g s, leaving g m r = g m+n. If m < r, then m of the g s cancel all the gs, leaving (g ) r m = g (r m) = g m+n. The argument is similar in the other two cases. The proof of (g m ) n = g mn also divides into a number of cases. When m or n is zero, both sides are the identity. When m and n are positive, then (g m ) n is the product of n terms, each the product of m factors g, giving the result g mn. The case m < 0 and n > 0 is similar with factors g instead. If m > 0 and n < 0, say n = r, then (g m ) n = (g m ) r is the product of r factors equal to (g m ) = (g ) m, so is the product of mr factors g ; thus it is equal to g mr = g mn. The last case is left to the reader. Finally, suppose that gh = hg and consider (gh) n. If n > 0, this is the product of n factors gh, which can be rearranged with all the gs at the beginning to give g n h n as required. If n < 0, say n = r, we have (gh) n = (gh) r = (hg) r = ((hg) r ) = (h r g r ) = (g r ) (h r ) = g r h r = g n h n. 2
3 (We use the fact that (xy) = y x here.) Finally, if n = 0, then both sides are the identity. sol3.3. We claim that, for any g G, the set ghg is a subgroup of G. [Apply the Subgroup Test: take two elements of ghg, say gxg and gyg, where x,y H. Then (gxg )(gyg ) ) = gxg gy g = g(xy )g ghg, since xy H.] Now the left coset gh of H is equal to (ghg )g, which is a right coset of the subgroup ghg (a) Lagrange s Theorem: if G contains an element of order 2, then 2 divides the order of G. (b) As suggested, let x,y,x 2,y 2,...,x m,y m be the elements of G which are not equal to their inverses, with the notation chosen so that xi = y i for i =,...,m; and let z,...,z r be the elements equal to their inverses. Then G = 2m + r. If G is even, then r is even. But the identity is equal to its inverse, so r. Hence r 2, and there is at least one nonidentity element z i, say z. Then z = z, so z 2 = ; since z, z has order We know that an element x Z m is a unit if and only if gcd(x,m) = (by Proposition 2.5). The number of units in Z m is thus equal to φ(m); in other words, φ(m) is the order of the group U(Z m ) of units of Z m. By Theorem 3.6(c), x φ(m) = in Z m, in other words, x φ(m) m. Suppose that m = p is a prime number. Then all the nonzero elements,..., p of Z p are units, since the only possible common divisor with p would be p itself, and none of these are divisible by p. So φ(p) = p, and x p p if x p 0. Multiplying both sides by x we see that x p p x if x p 0. But this congruence holds also if x p 0; so it holds for all elements of Z p, in other words, all integers x satisfy x p p x. 3.9 We show first that the group G is isomorphic to the group G consisting of all transformations of F of the form θ a,b : x ax + b, where a,b F and a 0. Clearly for every matrix there is such a transformation, so the map is a bijection. We check the homomorphism property: while in other words, ( a b 0 )( ) c d = 0 ( ) ac ad + b, 0 θ a,b θ c,d : x (cx + d) a(cx + d) + b = acx + (ad + b), θ a,b θ c,d = θ ac,ad+b. Now suppose that F = Z 3. Then G is a group with 2 3 = 6 elements (since there are 2 choices for a and 3 for b), each element of which is a permutation of F. Since the symmetric group on F has only six elements, we must have G = S 3, and so G is isomorphic to S 3. 3
4 3.2 The fact that G/Z(G) is cyclic, generated by Z(G)g, means that G/Z(G) (the set of cosets of Z(G) in G) consists of all the powers (Z(G)g) i = Z(G)g i. So every coset has this form. Now every element h G lies in a unique coset of Z(G), of the form Z(G)g i for some i; thus h = zg i for some z Z(G). Take two elements h and h 2 of G; say h = z g i and h 2 = z 2 g j for some z,z 2 Z(G) and i, j Z. Then h h 2 = z g i z 2 g j = z z 2 g i+ j = z 2 z g i+ j = z 2 g j z g i = h 2 h, where for the second inequality we use the fact that z 2 commutes with g i ; for the third, the fact that z and z 2 commute; and for the fourth, the fact that z commutes with g i. Thus h and h 2 commute. Since they were arbitrary elements of G, we see that G is abelian, and indeed G = Z(G) The group S 3 has order 6. By Lagrange s Theorem, any subgroup has order, 2, 3 or 6. The only subgroup of order is {}, and the only subgroup of order 6 is S 3. So we have to look for subgroups of orders 2 and 3. Note that, as well as the identity, S 3 contains two elements of order 3 (viz. (,2,3) and (,3,2)), which are inverses of each other, and three elements of order 2 (viz. (,2), (,3) and (2,3)). Again by Lagrange, if H is a subgroup of order 3, then every element of H must have order or 3. There are only three such elements altogether, namely, (,2,3) and (,3,2); so these form the only possible such subgroup. But this set is indeed a subgroup. So there is one subgroup of order 3. If K is a subgroup of order 2, then any element of K has order or 2; K must contain the identity, so must consist of the identity and a single element of order 2. Thus there are three possibilities for K, and it is routine to check that each of them is a subgroup. So there are altogether six subgroups of S Note that, since N is a normal subgroup of G, for any elements n N and g G, there exists n N such that gn = n g. (This is because gn lies in the left coset gn, which equals the right coset Ng.) We apply the first subgroup test. Take two elements of NH, say n h and n 2 h 2, where n,n 2 N and h,h 2 H. Their product is (n h ) (n 2 h 2 ) = n (h n 2 )h 2 = n (n h )h 2 = (n n )(h h 2 ) NH, where n is some element of N. Take an element nh NH. Its inverse is for some n N. So NH is a subgroup of G. (nh) = h n = n h NH, (a) True. If also H is a normal subgroup of G, take any nh NH and g G; we have g(nh) = (gn)h = (n g)h = n (gh) = n (h g) = (n h )g, 4
5 so left and right cosets of NH are equal. (b) False. Let G = S 3, N = A 3 = {,(,2,3),(,3,2)} and H = {,(,2)}. (See Exercise 3.23.) Then NH = G, so NH is certainly a normal subgroup of G; but H is not a normal subgroup (a) Suppose that G is a group of finite order n which has just two conjugacy classes. One of these classes consists of the identity; so the other has size n. Now the size of a conjugacy class is the index of the centraliser of one of its elements, and so divides G (see Theorem 3.2); so n divides n. It follows that n divides n (n ) = ; so n =, and n = 2, G = C 2. The hint suggests using the class equation, which would say n + k =, where k is the order of the centraliser of a nonidentity element. This equation has only the solution n = n = 2. [Why?] (b) Show directly that the conjugacy classes in S 3 are {}, {(,2,3),(,3,2)}, and {(,2),(,3),(2,3)}. (Elements in different sets in this list have different orders and so cannot be conjugate; your job is to show that elements in the same set are conjugate.) The class equation in this case becomes =. (c) Suppose that G has three conjugacy classes, and has order n; let k and l be the orders of the centralisers of elements in the other two classes. Then n + k + l =. If three unit fractions sum to, then the largest of them is at least /3; so, without loss of generality, l = 2 or l = 3. If l = 3, then the only possibility is /3 + /3 + /3 =, so G = 3 (and the cyclic group of order 3 does indeed have three conjugacy classes). If l = 2, then /n + /k = /2, so k 4. We have two possible solutions; (n,k,l) = (4,4,2) and (n,k,l) = (6,3,2). So indeed G 6. But indeed the solution (4,4,2) is impossible, since a group of order 4 is abelian and so has four conjugacy classes. So there are just two finite groups with three conjugacy classes. (d) This will be a nonconstructive proof; that is, we will prove that the function exists without actually finding any information about it. As a harder exercise, you are encouraged to find estimates for the value of f (r). We start with the class equation for a group with r conjugacy classes: n + n n r =, where n,...,n r are the orders of centralisers of elements in the conjugacy classes. Note that the group order is the largest of the numbers n,...,n r (since the centraliser of the indentity is the whole group). So the result will follow if we can show that, given 5
6 r, this equation has only a finite number of solutions: then we can take f (r) to be the largest number appearing in any such solution. In order to prove this, we use induction on r, but we have to prove a more general statement: Lemma Given any positive integer r and rational number q, there are only finitely many rtuples (n,n 2,...,n r ) of natural numbers such that n + n n r = q. For q = this gives the desired conclusion. Proof The proof is by induction on r. For r =, the equation has one solution if q is the reciprocal of a natural number and none otherwise. Suppose that the lemma is true for r. Consider any solution of the equation with the given values of q and r. If n r is the smallest of the numbers n i, then /n r is the largest fraction, so it is at least as great as the average q/r; so n r r/q. Now for each possible value of n r in the range [,r/q], we have the equation + + = q, n n r n r and by the induction hypothesis, each of these equations has only finitely many solutions. So there are only finitely many solutions altogether. Remark Note how the argument forced us to prove a more general result; even though we are really interested in equations with righthand side equal to, we have to allow other values in order to use induction Construct a Cayley table for the four elements. 3.3 How many times does the element g i occur in row r of the table? If such an occurrence occurs in column s, then g r g s = g i, so g s = g r g i. So there is exactly one position in row r where g i appears. The argument for columns is similar (a) First proof: Let C 2 = {,g}, where g 2 =. The Cayley table for C 2 C 2 is (, ) (, g) (g, ) (g, g) (,) (,) (,g) (g,) (g,g) (,g) (,g) (,) (g,g) (g,) (g,) (g,) (g,g) (,) (,g) (g,g) (g,g) (g,) (,g) (,) which is easily matched up with the Cayley table for V 4. 6
7 Second proof: C 2 C 2 is a group of order 4 which is easily seen to have no elements of order 4; so it is not isomorphic to C 4, and must be isomorphic to V 4 (see p.32). (b) First proof: Let C 2 be as in (a), and C 3 = {,h,h 2 }, with h 3 =. The order of (g,h) C 2 C 3 must divide 6; it is not 2 (since (g,h) 2 = (,h 2 )), and is not 3 (since (g,h) 3 = (g,)); so it has order 6, and generates the cyclic group. Second proof: By the preceding exercise, C 2 C 3 is an abelian group of order 6. Now apply the classification on p.33. See p.34. C 8 is obtained if there is an element of order 8, and C 2 C 2 C 2 if there is no element of order 4. We obtain C 2 C 4 in the case where b 2 = and ba = ab, and also in the case where b 2 = a 2 and ba = ab (a) As in the first proof in Exercise 3.33(b), let C p = g and C q = h. Then the element (g,h) of C p C q has order dividing pq, but not p or q; so its order is pq, and the group is cyclic. (b) In C p C p, every element has order or p; so the group is not cyclic For convenience we represent the elements of G by permutations, as on p.39. Check that z = (,3)(2,4) commutes with all elements of G. (In fact, we know that every element of G can be written in the form a i b j, where a = (,2,3,4) and b = (,4)(2,3) (see the analysis on p.34); so it is enough to show that z commutes with a and b.) So the subgroup Z = {,z} is contained in Z(G). If Z(G) were larger than Z, then its order would be 4 or 8, so that G/Z(G) would have order or 2, and would be cyclic; by Exercise 3.2, G would be abelian, which it is not. So Z(G) = Z. (You can check this directly by showing that for any element g G apart from and z, there is an element h G which does not commute with g.) Now the only elements of order 4 in G are a and a 3, and a 3 = za, so Za = Za 3, and (Za) 2 = Z. Thus the factor group G/Z is a group of order 4 in which no element has order greater than 2, and is necessarily the Klein group. (More simply, invoke Exercise 3.2 again to see that G/Z(G) cannot be cyclic.) 3.39 Take m M and n N. Consider the element g = m n mn, the socalled commutator of m and n. Writing it as m (n mn), we see that it is the inverse of m times a conjugate of m; both of these lie in M, so g M. Similarly, writing it as g = (m n m)n, we see that G N. Since M N = {}, we see that g =, so that mn = nm. Since G = MN, every element of G can be written in the form g = mn for m M and n N. Suppose we have another representation, g = m n with m M and n N. Then mn = m n. Multiplying this equation on the left by (m ) and on the right by n, we obtain (m ) m = n n. The lefthand expression lies in M and the righthand one in N; so both are the identity, giving m = m and n = n. Now define a map θ from G to M N by θ : mn G (m,n) M N. By what we just proved, this map is welldefined. Clearly it is onetoone and onto. Also, if mn,m n G (with m,m M and n,n N), then (mn)(m n ) = (mm )(nn ) by 7
8 what we showed in the first paragraph; so ((mn)(m n ))θ = ((mm )(nn ))θ = (mm,nn ), (mn)θ (m n )θ = (m,n)(m,n ) = (mm,nn ), where the last equation is the definition of the group operation in M N. So θ is an isomorphism. Let N be the rotation group of the cube and M = {±I}. Since every rotation has determinant, we have M N = {I}. Also, G = 48, N = 24 and M = 2, so MN = G and MN = G. Moreover, both M and N are normal subgroups (N because it has index 2, and M because its elements commute with everything so it is in Z(G)). So the conditions of the first part of the exercise are satisfied, and we conclude that G = M N. 3.4 The first part of this exercise is obvious from playing with a model. It could of course be proved by coordinate geometry but I do not expect you to do this! Clearly any rotation maps a frame to a frame, so induces a permutation on the set of five frames. Could a rotation fix every frame? Again, a few moments playing with a model shows that only the identity does so. So the map from the rotation group G to the group S 5 of permutations of the frames is a onetoone homomorphism. Its image is a subgroup of S 5 having order 60, so a normal subgroup, necessarily A 5. 8
Definitions, Theorems and Exercises. Abstract Algebra Math 332. Ethan D. Bloch
Definitions, Theorems and Exercises Abstract Algebra Math 332 Ethan D. Bloch December 26, 2013 ii Contents 1 Binary Operations 3 1.1 Binary Operations............................... 4 1.2 Isomorphic Binary
More informationA. (Groups of order 8.) (a) Which of the five groups G (as specified in the question) have the following property: G has a normal subgroup N such that
MATH 402A  Solutions for the suggested problems. A. (Groups of order 8. (a Which of the five groups G (as specified in the question have the following property: G has a normal subgroup N such that N =
More informationBasic Definitions: Group, subgroup, order of a group, order of an element, Abelian, center, centralizer, identity, inverse, closed.
Math 546 Review Exam 2 NOTE: An (*) at the end of a line indicates that you will not be asked for the proof of that specific item on the exam But you should still understand the idea and be able to apply
More informationSelected exercises from Abstract Algebra by Dummit and Foote (3rd edition).
Selected exercises from Abstract Algebra by Dummit Foote (3rd edition). Bryan Félix Abril 12, 2017 Section 4.1 Exercise 1. Let G act on the set A. Prove that if a, b A b = ga for some g G, then G b = gg
More informationDefinitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations
Page 1 Definitions Tuesday, May 8, 2018 12:23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of
More informationCONSEQUENCES OF THE SYLOW THEOREMS
CONSEQUENCES OF THE SYLOW THEOREMS KEITH CONRAD For a group theorist, Sylow s Theorem is such a basic tool, and so fundamental, that it is used almost without thinking, like breathing. Geoff Robinson 1.
More informationThe Symmetric Groups
Chapter 7 The Symmetric Groups 7. Introduction In the investigation of finite groups the symmetric groups play an important role. Often we are able to achieve a better understanding of a group if we can
More informationZachary Scherr Math 370 HW 7 Solutions
1 Book Problems 1. 2.7.4b Solution: Let U 1 {u 1 u U} and let S U U 1. Then (U) is the set of all elements of G which are finite products of elements of S. We are told that for any u U and g G we have
More informationφ(xy) = (xy) n = x n y n = φ(x)φ(y)
Groups 1. (Algebra Comp S03) Let A, B and C be normal subgroups of a group G with A B. If A C = B C and AC = BC then prove that A = B. Let b B. Since b = b1 BC = AC, there are a A and c C such that b =
More informationits image and kernel. A subgroup of a group G is a nonempty subset K of G such that k 1 k 1
10 Chapter 1 Groups 1.1 Isomorphism theorems Throughout the chapter, we ll be studying the category of groups. Let G, H be groups. Recall that a homomorphism f : G H means a function such that f(g 1 g
More informationNormal Subgroups and Factor Groups
Normal Subgroups and Factor Groups Subject: Mathematics Course Developer: Harshdeep Singh Department/ College: Assistant Professor, Department of Mathematics, Sri Venkateswara College, University of Delhi
More informationMath 210A: Algebra, Homework 5
Math 210A: Algebra, Homework 5 Ian Coley November 5, 2013 Problem 1. Prove that two elements σ and τ in S n are conjugate if and only if type σ = type τ. Suppose first that σ and τ are cycles. Suppose
More informationMA441: Algebraic Structures I. Lecture 26
MA441: Algebraic Structures I Lecture 26 10 December 2003 1 (page 179) Example 13: A 4 has no subgroup of order 6. BWOC, suppose H < A 4 has order 6. Then H A 4, since it has index 2. Thus A 4 /H has order
More informationAutomorphism Groups Definition. An automorphism of a group G is an isomorphism G G. The set of automorphisms of G is denoted Aut G.
Automorphism Groups 992012 Definition. An automorphism of a group G is an isomorphism G G. The set of automorphisms of G is denoted Aut G. Example. The identity map id : G G is an automorphism. Example.
More informationExercises on chapter 1
Exercises on chapter 1 1. Let G be a group and H and K be subgroups. Let HK = {hk h H, k K}. (i) Prove that HK is a subgroup of G if and only if HK = KH. (ii) If either H or K is a normal subgroup of G
More information5 Group theory. 5.1 Binary operations
5 Group theory This section is an introduction to abstract algebra. This is a very useful and important subject for those of you who will continue to study pure mathematics. 5.1 Binary operations 5.1.1
More informationMath 250A, Fall 2004 Problems due October 5, 2004 The problems this week were from Lang s Algebra, Chapter I.
Math 250A, Fall 2004 Problems due October 5, 2004 The problems this week were from Lang s Algebra, Chapter I. 24. We basically know already that groups of order p 2 are abelian. Indeed, pgroups have nontrivial
More informationSelected exercises from Abstract Algebra by Dummit and Foote (3rd edition).
Selected exercises from Abstract Algebra by Dummit and Foote (3rd edition). Bryan Félix Abril 12, 2017 Section 2.1 Exercise (6). Let G be an abelian group. Prove that T = {g G g < } is a subgroup of G.
More information6 Cosets & Factor Groups
6 Cosets & Factor Groups The course becomes markedly more abstract at this point. Our primary goal is to break apart a group into subsets such that the set of subsets inherits a natural group structure.
More informationFROM GROUPS TO GALOIS Amin Witno
WON Series in Discrete Mathematics and Modern Algebra Volume 6 FROM GROUPS TO GALOIS Amin Witno These notes 1 have been prepared for the students at Philadelphia University (Jordan) who are taking the
More informationINTRODUCTION TO THE GROUP THEORY
Lecture Notes on Structure of Algebra INTRODUCTION TO THE GROUP THEORY By : Drs. Antonius Cahya Prihandoko, M.App.Sc email: antoniuscp.fkip@unej.ac.id Mathematics Education Study Program Faculty of Teacher
More informationCosets, factor groups, direct products, homomorphisms, isomorphisms
Cosets, factor groups, direct products, homomorphisms, isomorphisms Sergei Silvestrov Spring term 2011, Lecture 11 Contents of the lecture Cosets and the theorem of Lagrange. Direct products and finitely
More informationMATH HL OPTION  REVISION SETS, RELATIONS AND GROUPS Compiled by: Christos Nikolaidis
MATH HL OPTION  REVISION SETS, RELATIONS AND GROUPS Compiled by: Christos Nikolaidis PART B: GROUPS GROUPS 1. ab The binary operation a * b is defined by a * b = a+ b +. (a) Prove that * is associative.
More informationExtra exercises for algebra
Extra exercises for algebra These are extra exercises for the course algebra. They are meant for those students who tend to have already solved all the exercises at the beginning of the exercise session
More informationSchool of Mathematics and Statistics. MT5824 Topics in Groups. Problem Sheet I: Revision and ReActivation
MRQ 2009 School of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet I: Revision and ReActivation 1. Let H and K be subgroups of a group G. Define HK = {hk h H, k K }. (a) Show that HK
More informationGroups Subgroups Normal subgroups Quotient groups Homomorphisms Cyclic groups Permutation groups Cayley s theorem Class equations Sylow theorems
Group Theory Groups Subgroups Normal subgroups Quotient groups Homomorphisms Cyclic groups Permutation groups Cayley s theorem Class equations Sylow theorems Groups Definition : A nonempty set ( G,*)
More information0 Sets and Induction. Sets
0 Sets and Induction Sets A set is an unordered collection of objects, called elements or members of the set. A set is said to contain its elements. We write a A to denote that a is an element of the set
More informationAlgebra Exercises in group theory
Algebra 3 2010 Exercises in group theory February 2010 Exercise 1*: Discuss the Exercises in the sections 1.11.3 in Chapter I of the notes. Exercise 2: Show that an infinite group G has to contain a nontrivial
More informationGroups and Symmetries
Groups and Symmetries Definition: Symmetry A symmetry of a shape is a rigid motion that takes vertices to vertices, edges to edges. Note: A rigid motion preserves angles and distances. Definition: Group
More informationTheorems and Definitions in Group Theory
Theorems and Definitions in Group Theory Shunan Zhao Contents 1 Basics of a group 3 1.1 Basic Properties of Groups.......................... 3 1.2 Properties of Inverses............................. 3
More informationAlgebraic structures I
MTH5100 Assignment 110 Algebraic structures I For handing in on various dates January March 2011 1 FUNCTIONS. Say which of the following rules successfully define functions, giving reasons. For each one
More informationWritten Homework # 2 Solution
Math 516 Fall 2006 Radford Written Homework # 2 Solution 10/09/06 Let G be a nonempty set with binary operation. For nonempty subsets S, T G we define the product of the sets S and T by If S = {s} is
More informationGroups. Chapter 1. If ab = ba for all a, b G we call the group commutative.
Chapter 1 Groups A group G is a set of objects { a, b, c, } (not necessarily countable) together with a binary operation which associates with any ordered pair of elements a, b in G a third element ab
More informationCHAPTER 9. Normal Subgroups and Factor Groups. Normal Subgroups
Normal Subgroups CHAPTER 9 Normal Subgroups and Factor Groups If H apple G, we have seen situations where ah 6= Ha 8 a 2 G. Definition (Normal Subgroup). A subgroup H of a group G is a normal subgroup
More informationCourse 311: Abstract Algebra Academic year
Course 311: Abstract Algebra Academic year 200708 D. R. Wilkins Copyright c David R. Wilkins 1997 2007 Contents 1 Topics in Group Theory 1 1.1 Groups............................... 1 1.2 Examples of Groups.......................
More informationB Sc MATHEMATICS ABSTRACT ALGEBRA
UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION B Sc MATHEMATICS (0 Admission Onwards) V Semester Core Course ABSTRACT ALGEBRA QUESTION BANK () Which of the following defines a binary operation on Z
More information2 Lecture 2: Logical statements and proof by contradiction Lecture 10: More on Permutations, Group Homomorphisms 31
Contents 1 Lecture 1: Introduction 2 2 Lecture 2: Logical statements and proof by contradiction 7 3 Lecture 3: Induction and WellOrdering Principle 11 4 Lecture 4: Definition of a Group and examples 15
More informationDMATH Algebra I HS 2013 Prof. Brent Doran. Solution 3. Modular arithmetic, quotients, product groups
DMATH Algebra I HS 2013 Prof. Brent Doran Solution 3 Modular arithmetic, quotients, product groups 1. Show that the functions f = 1/x, g = (x 1)/x generate a group of functions, the law of composition
More informationElements of solution for Homework 5
Elements of solution for Homework 5 General remarks How to use the First Isomorphism Theorem A standard way to prove statements of the form G/H is isomorphic to Γ is to construct a homomorphism ϕ : G Γ
More informationMODEL ANSWERS TO THE FIFTH HOMEWORK
MODEL ANSWERS TO THE FIFTH HOMEWORK 1. Chapter 3, Section 5: 1 (a) Yes. Given a and b Z, φ(ab) = [ab] = [a][b] = φ(a)φ(b). This map is clearly surjective but not injective. Indeed the kernel is easily
More informationExamples: The (left or right) cosets of the subgroup H = 11 in U(30) = {1, 7, 11, 13, 17, 19, 23, 29} are
Cosets Let H be a subset of the group G. (Usually, H is chosen to be a subgroup of G.) If a G, then we denote by ah the subset {ah h H}, the left coset of H containing a. Similarly, Ha = {ha h H} is the
More informationSolutions to oddnumbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 2
Solutions to oddnumbered exercises Peter J Cameron, Introduction to Algebra, Chapter 1 The answers are a No; b No; c Yes; d Yes; e No; f Yes; g Yes; h No; i Yes; j No a No: The inverse law for addition
More information1. Group Theory Permutations.
1.1. Permutations. 1. Group Theory Problem 1.1. Let G be a subgroup of S n of index 2. Show that G = A n. Problem 1.2. Find two elements of S 7 that have the same order but are not conjugate. Let π S 7
More informationCHAPTER III NORMAL SERIES
CHAPTER III NORMAL SERIES 1. Normal Series A group is called simple if it has no nontrivial, proper, normal subgroups. The only abelian simple groups are cyclic groups of prime order, but some authors
More informationMath 120. Groups and Rings Midterm Exam (November 8, 2017) 2 Hours
Math 120. Groups and Rings Midterm Exam (November 8, 2017) 2 Hours Name: Please read the questions carefully. You will not be given partial credit on the basis of having misunderstood a question, and please
More informationSF2729 GROUPS AND RINGS LECTURE NOTES
SF2729 GROUPS AND RINGS LECTURE NOTES 20110301 MATS BOIJ 6. THE SIXTH LECTURE  GROUP ACTIONS In the sixth lecture we study what happens when groups acts on sets. 1 Recall that we have already when looking
More informationA connection between number theory and linear algebra
A connection between number theory and linear algebra Mark Steinberger Contents 1. Some basics 1 2. Rational canonical form 2 3. Prime factorization in F[x] 4 4. Units and order 5 5. Finite fields 7 6.
More informationYale University Department of Mathematics Math 350 Introduction to Abstract Algebra Fall Midterm Exam Review Solutions
Yale University Department of Mathematics Math 350 Introduction to Abstract Algebra Fall 2015 Midterm Exam Review Solutions Practice exam questions: 1. Let V 1 R 2 be the subset of all vectors whose slope
More informationPhysics 251 Solution Set 1 Spring 2017
Physics 5 Solution Set Spring 07. Consider the set R consisting of pairs of real numbers. For (x,y) R, define scalar multiplication by: c(x,y) (cx,cy) for any real number c, and define vector addition
More information1.5 Applications Of The Sylow Theorems
14 CHAPTER1. GROUP THEORY 8. The Sylow theorems are about subgroups whose order is a power of a prime p. Here is a result about subgroups of index p. Let H be a subgroup of the finite group G, and assume
More informationMath 345 Sp 07 Day 7. b. Prove that the image of a homomorphism is a subring.
Math 345 Sp 07 Day 7 1. Last time we proved: a. Prove that the kernel of a homomorphism is a subring. b. Prove that the image of a homomorphism is a subring. c. Let R and S be rings. Suppose R and S are
More informationPh.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018
Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018 Do 6 problems with at least 2 in each section. Group theory problems: (1) Suppose G is a group. The
More informationTHE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM  FALL SESSION ADVANCED ALGEBRA I.
THE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM  FALL SESSION 2006 110.401  ADVANCED ALGEBRA I. Examiner: Professor C. Consani Duration: take home final. No calculators allowed.
More informationMath 3140 Fall 2012 Assignment #3
Math 3140 Fall 2012 Assignment #3 Due Fri., Sept. 21. Remember to cite your sources, including the people you talk to. My solutions will repeatedly use the following proposition from class: Proposition
More informationSection III.15. FactorGroup Computations and Simple Groups
III.15 FactorGroup Computations 1 Section III.15. FactorGroup Computations and Simple Groups Note. In this section, we try to extract information about a group G by considering properties of the factor
More informationStab(t) = {h G h t = t} = {h G h (g s) = g s} = {h G (g 1 hg) s = s} = g{k G k s = s} g 1 = g Stab(s)g 1.
1. Group Theory II In this section we consider groups operating on sets. This is not particularly new. For example, the permutation group S n acts on the subset N n = {1, 2,...,n} of N. Also the group
More informationFall /29/18 Time Limit: 75 Minutes
Math 411: Abstract Algebra Fall 2018 Midterm 10/29/18 Time Limit: 75 Minutes Name (Print): Solutions JHUID: This exam contains 8 pages (including this cover page) and 6 problems. Check to see if any pages
More informationCHAPTER 8: EXPLORING R
CHAPTER 8: EXPLORING R LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN In the previous chapter we discussed the need for a complete ordered field. The field Q is not complete, so we constructed
More informationAlgebra SEP Solutions
Algebra SEP Solutions 17 July 2017 1. (January 2017 problem 1) For example: (a) G = Z/4Z, N = Z/2Z. More generally, G = Z/p n Z, N = Z/pZ, p any prime number, n 2. Also G = Z, N = nz for any n 2, since
More informationMath 4320, Spring 2011
Math 4320, Spring 2011 Prelim 2 with solutions 1. For n =16, 17, 18, 19 or 20, express Z n (A product can have one or more factors.) as a product of cyclic groups. Solution. For n = 16, G = Z n = {[1],
More informationChapter 3. Introducing Groups
Chapter 3 Introducing Groups We need a supermathematics in which the operations are as unknown as the quantities they operate on, and a supermathematician who does not know what he is doing when he performs
More informationx 2 = xn xn = x 2 N = N = 0
Potpourri. Spring 2010 Problem 2 Let G be a finite group with commutator subgroup G. Let N be the subgroup of G generated by the set {x 2 : x G}. Then N is a normal subgroup of G and N contains G. Proof.
More informationTeddy Einstein Math 4320
Teddy Einstein Math 4320 HW4 Solutions Problem 1: 2.92 An automorphism of a group G is an isomorphism G G. i. Prove that Aut G is a group under composition. Proof. Let f, g Aut G. Then f g is a bijective
More informationMATH 436 Notes: Cyclic groups and Invariant Subgroups.
MATH 436 Notes: Cyclic groups and Invariant Subgroups. Jonathan Pakianathan September 30, 2003 1 Cyclic Groups Now that we have enough basic tools, let us go back and study the structure of cyclic groups.
More informationAssigment 1. 1 a b. 0 1 c A B = (A B) (B A). 3. In each case, determine whether G is a group with the given operation.
1. Show that the set G = multiplication. Assigment 1 1 a b 0 1 c a, b, c R 0 0 1 is a group under matrix 2. Let U be a set and G = {A A U}. Show that G ia an abelian group under the operation defined by
More information1 Fields and vector spaces
1 Fields and vector spaces In this section we revise some algebraic preliminaries and establish notation. 1.1 Division rings and fields A division ring, or skew field, is a structure F with two binary
More information7 Semidirect product. Notes 7 Autumn Definition and properties
MTHM024/MTH74U Group Theory Notes 7 Autumn 20 7 Semidirect product 7. Definition and properties Let A be a normal subgroup of the group G. A complement for A in G is a subgroup H of G satisfying HA = G;
More informationDISCRETE MATH (A LITTLE) & BASIC GROUP THEORY  PART 3/3. Contents
DISCRETE MATH (A LITTLE) & BASIC GROUP THEORY  PART 3/3 T.K.SUBRAHMONIAN MOOTHATHU Contents 1. Cayley s Theorem 1 2. The permutation group S n 2 3. Center of a group, and centralizers 4 4. Group actions
More informationSolutions of exercise sheet 4
DMATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 4 The content of the marked exercises (*) should be known for the exam. 1. Prove the following two properties of groups: 1. Every
More informationA Little Beyond: Linear Algebra
A Little Beyond: Linear Algebra Akshay Tiwary March 6, 2016 Any suggestions, questions and remarks are welcome! 1 A little extra Linear Algebra 1. Show that any set of nonzero polynomials in [x], no two
More information* 8 Groups, with Appendix containing Rings and Fields.
* 8 Groups, with Appendix containing Rings and Fields Binary Operations Definition We say that is a binary operation on a set S if, and only if, a, b, a b S Implicit in this definition is the idea that
More informationNote that a unit is unique: 1 = 11 = 1. Examples: Nonnegative integers under addition; all integers under multiplication.
Algebra fact sheet An algebraic structure (such as group, ring, field, etc.) is a set with some operations and distinguished elements (such as 0, 1) satisfying some axioms. This is a fact sheet with definitions
More informationABSTRACT ALGEBRA 1, LECTURES NOTES 5: SUBGROUPS, CONJUGACY, NORMALITY, QUOTIENT GROUPS, AND EXTENSIONS.
ABSTRACT ALGEBRA 1, LECTURES NOTES 5: SUBGROUPS, CONJUGACY, NORMALITY, QUOTIENT GROUPS, AND EXTENSIONS. ANDREW SALCH 1. Subgroups, conjugacy, normality. I think you already know what a subgroup is: Definition
More informationMath 2070BC Term 2 Weeks 1 13 Lecture Notes
Math 2070BC 2017 18 Term 2 Weeks 1 13 Lecture Notes Keywords: group operation multiplication associative identity element inverse commutative abelian group Special Linear Group order infinite order cyclic
More informationSupplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.
Glossary 1 Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.23 Abelian Group. A group G, (or just G for short) is
More information2) e = e G G such that if a G 0 =0 G G such that if a G e a = a e = a. 0 +a = a+0 = a.
Chapter 2 Groups Groups are the central objects of algebra. In later chapters we will define rings and modules and see that they are special cases of groups. Also ring homomorphisms and module homomorphisms
More informationORDERS OF ELEMENTS IN A GROUP
ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since
More informationALGEBRA. 1. Some elementary number theory 1.1. Primes and divisibility. We denote the collection of integers
ALGEBRA CHRISTIAN REMLING 1. Some elementary number theory 1.1. Primes and divisibility. We denote the collection of integers by Z = {..., 2, 1, 0, 1,...}. Given a, b Z, we write a b if b = ac for some
More informationS11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 2 BASIC PROPERTIES
S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 2 BASIC PROPERTIES 1 Some Definitions For your convenience, we recall some of the definitions: A group G is called simple if it has
More informationRings. Chapter 1. Definition 1.2. A commutative ring R is a ring in which multiplication is commutative. That is, ab = ba for all a, b R.
Chapter 1 Rings We have spent the term studying groups. A group is a set with a binary operation that satisfies certain properties. But many algebraic structures such as R, Z, and Z n come with two binary
More informationCYCLICITY OF (Z/(p))
CYCLICITY OF (Z/(p)) KEITH CONRAD 1. Introduction For each prime p, the group (Z/(p)) is cyclic. We will give seven proofs of this fundamental result. A common feature of the proofs that (Z/(p)) is cyclic
More informationCONJUGATION IN A GROUP
CONJUGATION IN A GROUP KEITH CONRAD 1. Introduction A reflection across one line in the plane is, geometrically, just like a reflection across any other line. That is, while reflections across two different
More informationMath 581 Problem Set 9
Math 581 Prolem Set 9 1. Let m and n e relatively prime positive integers. (a) Prove that Z/mnZ = Z/mZ Z/nZ as RINGS. (Hint: First Isomorphism Theorem) Proof: Define ϕz Z/mZ Z/nZ y ϕ(x) = ([x] m, [x] n
More informationHomework #11 Solutions
Homework #11 Solutions p 166, #18 We start by counting the elements in D m and D n, respectively, of order 2. If x D m and x 2 then either x is a flip or x is a rotation of order 2. The subgroup of rotations
More information1 First Theme: Sums of Squares
I will try to organize the work of this semester around several classical questions. The first is, When is a prime p the sum of two squares? The question was raised by Fermat who gave the correct answer
More information1 Chapter 6  Exercise 1.8.cf
1 CHAPTER 6  EXERCISE 1.8.CF 1 1 Chapter 6  Exercise 1.8.cf Determine 1 The Class Equation of the dihedral group D 5. Note first that D 5 = 10 = 5 2. Hence every conjugacy class will have order 1, 2
More informationAlgebraI, Fall Solutions to Midterm #1
AlgebraI, Fall 2018. Solutions to Midterm #1 1. Let G be a group, H, K subgroups of G and a, b G. (a) (6 pts) Suppose that ah = bk. Prove that H = K. Solution: (a) Multiplying both sides by b 1 on the
More informationGroup Theory
Group Theory 2014 2015 Solutions to the exam of 4 November 2014 13 November 2014 Question 1 (a) For every number n in the set {1, 2,..., 2013} there is exactly one transposition (n n + 1) in σ, so σ is
More informationMath 594. Solutions 5
Math 594. Solutions 5 Book problems 6.1: 7. Prove that subgroups and quotient groups of nilpotent groups are nilpotent (your proof should work for infinite groups). Give an example of a group G which possesses
More informationPRACTICE FINAL MATH , MIT, SPRING 13. You have three hours. This test is closed book, closed notes, no calculators.
PRACTICE FINAL MATH 18.703, MIT, SPRING 13 You have three hours. This test is closed book, closed notes, no calculators. There are 11 problems, and the total number of points is 180. Show all your work.
More informationMATH 28A MIDTERM 2 INSTRUCTOR: HAROLD SULTAN
NAME: MATH 28A MIDTERM 2 INSTRUCTOR: HAROLD SULTAN 1. INSTRUCTIONS (1) Timing: You have 80 minutes for this midterm. (2) Partial Credit will be awarded. Please show your work and provide full solutions,
More informationINVERSE LIMITS AND PROFINITE GROUPS
INVERSE LIMITS AND PROFINITE GROUPS BRIAN OSSERMAN We discuss the inverse limit construction, and consider the special case of inverse limits of finite groups, which should best be considered as topological
More informationHomework #5 Solutions
Homework #5 Solutions p 83, #16. In order to find a chain a 1 a 2 a n of subgroups of Z 240 with n as large as possible, we start at the top with a n = 1 so that a n = Z 240. In general, given a i we will
More informationDefinition 1.2. Let G be an arbitrary group and g an element of G.
1. Group Theory I Section 2.2 We list some consequences of Lagrange s Theorem for exponents and orders of elements which will be used later, especially in 2.6. Definition 1.1. Let G be an arbitrary group
More information( ) 3 = ab 3 a!1. ( ) 3 = aba!1 a ( ) = 4 " 5 3 " 4 = ( )! 2 3 ( ) =! 5 4. Math 546 Problem Set 15
Math 546 Problem Set 15 1. Let G be a finite group. (a). Suppose that H is a subgroup of G and o(h) = 4. Suppose that K is a subgroup of G and o(k) = 5. What is H! K (and why)? Solution: H! K = {e} since
More informationCosets and Lagrange s theorem
Cosets and Lagrange s theorem These are notes on cosets and Lagrange s theorem some of which may already have been lecturer. There are some questions for you included in the text. You should write the
More informationMathematics 331 Solutions to Some Review Problems for Exam a = c = 3 2 1
Mathematics 331 Solutions to Some Review Problems for Exam 2 1. Write out all the even permutations in S 3. Solution. The six elements of S 3 are a =, b = 1 3 2 2 1 3 c =, d = 3 2 1 2 3 1 e =, f = 3 1
More informationTHE SYLOW THEOREMS AND THEIR APPLICATIONS
THE SYLOW THEOREMS AND THEIR APPLICATIONS AMIN IDELHAJ Abstract. This paper begins with an introduction into the concept of group actions, along with the associated notions of orbits and stabilizers, culminating
More informationPseudo Sylow numbers
Pseudo Sylow numbers Benjamin Sambale May 16, 2018 Abstract One part of Sylow s famous theorem in group theory states that the number of Sylow p subgroups of a finite group is always congruent to 1 modulo
More informationNOTES ON FINITE FIELDS
NOTES ON FINITE FIELDS AARON LANDESMAN CONTENTS 1. Introduction to finite fields 2 2. Definition and constructions of fields 3 2.1. The definition of a field 3 2.2. Constructing field extensions by adjoining
More information